Calculate the mass percent of Na3PO4.H2O in the sample

Question

A 1.5 g sample of a mixture of solid Na3PO4.H2O mixed with NaCl, was dissolved in water, reacted with aqueous 1 M SrCl2 and 0.52g Sr3(PO4)2 precipitate was collected and dried to a constant mass. If the reagent tests demonstrated that Na3PO4.12H2O was the limiting reactant in the precipitation reaction, calculate the mass percent of Na3PO4.12H2O in the original sample.

Answer 1

2Na₃PO₄-12H₂O + 3SrCl2 ==> 6NaCl(aq) + Sr₃(PO₄)₂(s) + 12H₂O .. balanced equation for reaction

molar mass Na₃PO₄-12H₂O = 380 g / mol

molar mass Sr₃(PO₄)₂ = 453 g / mol

Calculate grams of Na₃PO₄-12H₂O originally present:

0.52 g Sr₃(PO₄)₂ x 1 mol Sr₃(PO₄)₂ / 453 g x 2mol Na₃PO₄-12H₂O / mol Sr₃(PO₄)₂ = 0.002296 mols Na₃PO₄-12H₂O x 380 g Na₃PO₄-12H₂O / mol = 0.872 g Na₃PO₄-12H₂O originally present

Calculate mass percent of Na₃PO₄-12H₂O in original sample:

0.872 g Na₃PO₄-12H₂O / 1.5 g (x100%) = 58% (2 sig. figs.)